The probability of sum $x+y$ to be greater than $20$

Question

The variable $x$ takes a value between $0$ and $10$ with uniform probability distribution.The variable $y$ takes a value between $0$ and $20$ with uniform probability distribution. The probability of the sum of variables $(x+y)$ being greater than $20$ is

a) $0.5\ \quad $ b) $0\ \quad $ c) $0.25\ \quad $ d) $0.33\ \quad $

My try:

for uniform probability distribution, $f(x)=\frac{1}{10-0}=\frac{1}{10}$

$f(y)=\frac{1}{20-0}=\frac{1}{20}$

$f(x+y) =\frac{1}{\infty-20}=0$

$\therefore P(x+y>20)=\int_0^{\infty}f(t)dt=0$

I am not sure if i am right. Please correct me if i am wrong. please help me I am weak in probability problems.

Answer 1

First, we have $$ P[x\gt\alpha]=\frac{10-\alpha}{10}[0\le\alpha\le10] $$ Therefore, $$ \begin{align} \frac1{20}\int_0^{20}P[x\gt20-y]\,\mathrm{d}y &=\frac1{20}\int_{10}^{20}\frac{y-10}{10}\,\mathrm{d}y\\ &=\frac1{200}\int_0^{10}y\,\mathrm{d}y\\[3pt] &=\frac14 \end{align} $$

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We can also notice that $(x,y)$ is uniformly distributed over a $10\times20$ rectangle and the triangle over which $x+y\gt20$ is one quarter of that rectangle.